The goals of this experiment are as follows:

• to study the principle of operation of a diffraction grating spectrometer;
• to test the Balmer relation for the hydrogen atom;
• to determination of the Rydberg constant, using the hydrogen data and the Balmer formula; and
• to determine the mass of the deuteron by measuring the isotope shift between the corresponding wavelengths of hydrogen and deuterium.

In the 1800s, it was discovered that heated elements emitted light of specific colors depending on the element being observed. The set of colors for each element is referred to as it spectrum. The hydrogen spectrum is the simplest and reveals a pattern which first suggested a model that atoms had discrete or quantized energy levels_._ The theoretical models formulated to explain this phenomenon led to the stunning breakthroughs of quantum mechanics. One aspect of the model was the empirical Balmer formula you will be studying in this experiment.

Deuterium, or heavy hydrogen, was discovered by Harold Urey in 1931. Since deuterium’s atomic structure is like hydrogen, it has a very similar spectrum. The nucleus of hydrogen is simply a proton. We now know that he nucleus of deuterium is a deuteron, a combination of a proton and a neutron. The mass difference of the nuclei of hydrogen and deuterium has the effect of shifting all the spectral lines of deuterium relative to those of hydrogen. Measuring that small difference gives us a measurement of the mass of the deuteron. You will do this in the lab!

Figure 1 shows a schematic diagram of the grating spectrometer.  The light source whose spectrum we wish to observe is positioned at the entrance slit, whose width can be adjusted. The collimator projects a parallel beam onto the grating. The telescope and cross-hair eyepiece is used to view the diffracted light, while the divided circle and vernier scale allow accurate measurements of angles. Our goal here is to understand the relationship among the spacing of the tiny grooves in the grating, the angle through which the light of different colors (wavelengths) is diffracted, and the wavelength diffracted into that angle. This idea will become clearer!

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Figure 1: Grating spectrometer.

Preliminary adjustments require that the telescope be properly focused, and the grating accurately positioned so that it is normal to the collimator.

At one end of the collimator is a slit illuminated by the light source. The collimator lens collects the light from the slit and forms a parallel beam directed toward the grating. The slit width may be adjusted by turning the slit width adjustment knob.

The grating can be rotated and locked by means of knobs on the spectrometer base. The grating equation for normal incidence is given by

nλ=dsinθnλ=dsinθ { $n\lambda = d\sin\theta$ ,

(1)

where nn { $n$  is the diffraction order number, dd { $d$ the grating spacing, and θθ { $\theta$  the diffraction angle, measured with respect to the grating normal. Refer to Jenkins and White, Fundamentals of Optics, or other books for a more detailed treatment of this subject.

The light exiting the grating arrives at the telescope in parallel bundles, each color arriving at a different angle.  Since the rays in each bundle are parallel, the image of the collimator slit is at infinity and the telescope, when focused at infinity, will form a sharp image of the slit at the telescope crosshairs.

Fig. 2 shows the spectrometer vernier angle scale. In order to read the angular position of the telescope on the divided circle, proceed as follows:

• Read down from the zero mark on the vernier scale to the next smaller line on the degree scale. In the example of Fig. 2 the result would be 20 ½ degrees or 20 degrees, 30 minutes. 
• Next, find the line on the vernier scale which best coincides with a line on the degree scale. In the example, the best alignment is at 15 minutes. 
• Finally, add the two readings:  20 degrees 30 minutes + 15 minutes = 20 degrees, 45 minutes.

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Figure 2: An example of a vernier scale reading 20 degrees, 45 minutes.

Note that there are two angle scales on the spectrometer located 180 degrees apart. They are provided for additional precision. Since it is impossible to machine the divided circle and grating table mount to be perfectly concentric, any eccentricity will yield an angle reading too large on one scale and too small on the other. Therefore, read both and average the results.

1. Slide the telescope eyepiece in or out to focus the cross hairs. - Place a low intensity mercury lamp at the entrance slit of the collimator. - Move the telescope so that it is co-linear with the collimator, putting the cross-hair intersection point very near the stationary edge of the slit. - Examine the locking and slow motion knobs for the grating table and telescope. Note that the slow motion controls function properly only when the locking knobs are tightened. Lock the telescope. - Position the grating holder in the central hole of the spectrometer table with the grating facing the collimator. Adjust the grating height. -  Turn the collimator focus knob until the pencil mark on the collimator tube is just visible. - Read the vernier scale for the _θ_0 position. - To take the data, move only the telescope.

In order to understand how the grating spectrometer works, let’s start by looking through the spectrometer you already adjusted. Place a low intensity mercury lamp at the entrance slit of the collimator. Being careful not to move the grating, swing the telescope left and right while looking through it.

Question 1: Make a sketch of what you see on both sides of the degree position. Label the colors you see, and interpret any repeated color patterns using Eq. (1).

Now, make a careful measurement of the angle for, say, the green line. Be sure to use the vernier scale to obtain the greatest possible precision.

Question 2: At what angles does the green line appear? (Does it appear at more that one angle?)

Question 3: Use the known value of the green line of Hg (_λ _= 546.1 nm) to calculate the distance between the grooves on the grating. (This is d in Eq. (1).)

Now that you understand how a grating spectrometer works, we will record and analyze the Balmer spectrum of Hydrogen. The equation for the Balmer spectrum is

1λ=R(122−1n2)1λ=R(122−1n2) { $\dfrac{1}{\lambda} = R\left(\dfrac{1}{2^2} - \dfrac{1}{n^2}\right)$

(2)

where n = 3, 4, 5 and 6 for the four readily visible spectral lines of the Balmer series.

In the Bohr model, assuming that the nuclear mass is much greater than the orbiting electron mass, the Rydberg constant may be derived and is found to be

R=e48ϵ20h3cmeR=e48ϵ20h3cme { $R = \dfrac{e^4}{8\epsilon_0^2h^3c}m_e$

(3)

where RR { $R$  is the Rydberg constant ee { $e$  is the electron charge, ϵ0ϵ0 { $\epsilon_0$  is the permittivity of free space, hh { $h$  is Planck's constant, cc { $c$  is the speed of light, and meme { $m_e$  is the rest mass of the electron. Once the spectrometer is properly calibrated, replace the mercury lamp with the hydrogen discharge tube.  As accurately as possible measure the wavelengths of the 4 visible lines in the hydrogen Balmer series for the 1st and second order diffraction on both sides of the grating.  You will notice that the lines on one side of the grating are brighter than the other side.  For the dimmer lines you may not be able to see all of the wavelengths, just measure what you can find.

You can use the multiple measurements of each line to help estimate an uncertainty in your data.

Question 4: Obtain a value for the Rydberg constant R from your data. 

Question 5: Compare your experimental value of R to that predicted by Eq. (3). 

Values for the physical constants needed may be found on the inside back cover of this manual.

In reality, the Bohr model of the hydrogen atom is not quite correct; strictly speaking, the electron does not orbit the “stationary” proton. Rather, both the electron and proton orbit their combined center of mass. The proper treatment of such a system requires that the use of reduced mass in place of the electron mass in Eq.(3). The reduced mass for hydrogen (proton-electron system) is

μH=mpmemp+meμH=mpmemp+me { $\mu_H = \dfrac{m_pm_e}{m_p+m_e}$

(4)

where meme { $m_e$  and mpmp { $m_p$  denote the masses of the electron and proton, respectively. Similarly, the reduced mass for deuterium (deuteron-electron system) is

μH=mdmemd+meμH=mdmemd+me { $\mu_H = \dfrac{m_d m_e}{m_d+m_e}$

(5)

where mdmd { $m_d$  is the mass of the deuteron. Therefore, a more precise form of the Balmer formula, Eq. (2), applied to the hydrogen atom becomes,

1λH=e48ϵ20h3cμH(122−1n2)1λH=e48ϵ20h3cμH(122−1n2) { $\dfrac{1}{\lambda_H} = \dfrac{e^4}{8\epsilon_0^2h^3c}\mu_H\left(\dfrac{1}{2^2} - \dfrac{1}{n^2}\right)$

(6)

where we have replaced the electron mass with the hydrogen reduced mass. For simplicity, we define

K≡e48ϵ20h3cK≡e48ϵ20h3c { $K \equiv \dfrac{e^4}{8\epsilon_0^2h^3c}$

(7a)

and

A≡(122−1n2)A≡(122−1n2) { $A \equiv \left( \dfrac{1}{2^2} - \dfrac{1}{n^2}\right)$

(7b)

so that the Balmer formula for hydrogen may be written as

1λH=KAμH { $\dfrac{1}{\lambda_H} = KA\mu_H$

(8)

and the Balmer formula for deuterium may be written as

1λD=KAμD { $\dfrac{1}{\lambda_D} = KA\mu_D$

(9)

Thus, any specific transition will produce slightly different wavelengths for different isotopes. This phenomenon is called the isotope shift of the atomic spectra, which led Harold Urey to the discovery of deuterium in 1931, and the Nobel Prize in 1934.

Question 6: By inverting Eqs. (8) and (9), taking the difference, and further manipulating the result, show that the following relation holds between the deuteron mass and the isotope shift between hydrogen and deuterium:

md=mpKAmp(λD−λH)+1 { $m_d = \dfrac{m_p}{KAm_p(\lambda_D-\lambda_H)+1}$

(10)

The goal of this experiment is to determine the mass of the deuteron from the measured wavelength shifts between corresponding spectral lines of hydrogen and deuterium.

You will be given a gas discharge tube containing a mixture of hydrogen and deuterium in such proportions as to make the isotope shift more easily detectable. Because the isotope shift is very small, – a fraction of a nanometer, – one needs a spectrometer with very high resolution in order to measure the wavelength difference. Neither the optical spectrometer nor the RSE spectrometer has the necessary resolution to make the measurement. 

There is a single station set up in the lab with an Ocean Optics HR-2000 high-resolution spectrometer (see Fig. 7). The HR-2000 is a computer controlled grating spectrometer capable of resolving 0.06 nm in the wavelength range 600 nm to 700 nm. A collimating lens coupled to a fiber optic cable brings light to the spectrometer.

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Figure 7: Setup for measuring isotope shift

To record a spectrum make sure that the HR-2000 is connected to a USB port on the computer and start the Oceanview software (see Fig. 8) by double-clicking its icon on the desktop. The program automatically detects the spectrometer and displays the data in real time.

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Figure 8: Screenshot of the OceanView software.

Place the collimating lens (in the ring stand) close to the hydrogen-deuterium discharge tube. The discharge tube has a short lifetime, so it is powered through a foot pedal for momentary use. While pressing the foot pedal aim the fiber optic collimating lens at the illuminated discharge tube. You should see a peak in the spectrum in the OceanView software near 650 nm. Adjust the position of the collimating lens so that the peak does not run off the top of the graph. Click the pause icon to capture the spectrum and release the foot pedal to turn off the discharge lamp.

Use the zoom functions to zoom in on the peak in the spectrum so that you can clearly see both lines. By clicking on the spectrum with the mouse you can read the wavelength associated with any point on the graph.

Question 7: Determine the wavelength of each peak in the spectrum. Estimate uncertainties.

Question 8: From the wavelength you measured use Eq. (10) to determine the deuteron mass. For sufficient precision, it is necessary to keep 5 significant figures, whenever possible.

Question 9: From the measurement uncertainties, estimate the uncertainty in the calculated deuteron mass.

Question 10: Is your result consistent with the notion that the deuteron is a proton and a neutron bound together?